Pseudo force as seen by an accelerating frame of refrence

Question

A question struck my mind when i was trying to solve the following problem,

I was able to solve it by just considering forces in the horizontal and vertical direction however the solution turned out to be very lengthy, i found a solution for it online which used psuedo/fictious forces to solve the problem, the solution was as follows

Suppose there is an observer on the wedge and the wedge is accelerating towards the left with acceleration $a$. Then the psuedo force on the block of mass $m$ is $ma$.

Balancing forces on $m$ normal to the plane we get $$N + ma\sin\theta = mg\cos\theta$$ Balancing forces on $M$ in $X$ direction $$F + N\sin\theta = F\cos\theta + Ma$$

Solving these two equations gives me the solution to the above question as follows

$$a = \frac{mg\sin\theta\cos\theta + F(1-\cos\theta)}{M+m\sin^2\theta}$$

This does match with my answer and with the answer in my textbook, however i do not understand why there would be a psuedo force on the wedge of mass $M$, amounting to $Ma$.

With respect to a observer on the wedge should there be no psuedo force on it?.

Any help would be highly appreciated!

Answer 1

As you calculated in the "regular" reference frame, there is a force acting on the wedge, causing it to accelerate. Changing to another frame of reference will not make that disappear, but rather will add another fictitious force to match the observed acceleration in the new frame of reference.

If you are "riding" on the wedge, the acceleration of the wedge should be $0$, so the force you calculated due to the interaction between the block and the wedge is just canceled by the new pseudo force.

To sum up, the fact that you went to the wedge's frame of reference doesn't mean it doesn't need a pseudo-force to explain it's position, nor does it mean that all of the other "real" forces you calculated just disappeared: on contrary! They both add up to give $0$ acceleration.

I hope this helped :)